timeDate

You might want to check out 'chron'.  This stores the time as days and
fractions of a day.

If you take the current date,
as.numeric(chron(dates.="11/23/2004"))
[1] 12745

you get the value above.  If you change this to millisecond, you get
as.numeric(chron(dates.="11/23/2004")) * 86400 * 1000
[1] 1.101168e+12

this value requires 46 bits and since a floating point number has 54 bits
of value, it should be enough to give you millisecond resolution and still
maintain the 'date'
__________________________________________________________
James Holtman        "What is the problem you are trying to solve?"
Executive Technical Consultant  --  Office of Technology, Convergys
james.holtman at convergys.com
+1 (513) 723-2929

                      Yasser El-Zein                                                                                                       
                      <abu3ammar at gmail.com>        To:       r-help at stat.math.ethz.ch                                                      
                      Sent by:                     cc:                                                                                     
                      r-help-bounces at stat.m        Subject:  Re: [R] timeDate                                                              
                      ath.ethz.ch                                                                                                          

                      11/23/2004 09:55                                                                                                     
                      Please respond to                                                                                                    
                      Yasser El-Zein                                                                                                       

I am looking for up to the millisecond resolution. Is there a package
that has that?

On Mon, 22 Nov 2004 21:48:20 +0000 (UTC), Gabor Grothendieck
Yasser El-Zein <abu3ammar <at> gmail.com> writes:

From the document it is apparent to me that I need as.POSIXct  (I have
a double representing the number of millis since 1/1/1970 and I need
to construct a datetime object). I see it showing how to construct the
time object from a string representing the time but now fro a double
of millis. Does anyone know hoe to do that?

If by millis you mean milliseconds (i.e. one thousandths of a second)
then POSIXct does not support that resolution, but if rounding to
seconds is ok then

  structure(round(x/1000), class = c("POSIXt", "POSIXct"))

should give it to you assuming x is the number of milliseconds.

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"james" == james holtman <james.holtman at convergys.com>
    on Tue, 23 Nov 2004 10:39:04 -0500 writes:
james> You might want to check out 'chron'.  This stores the
    james> time as days and fractions of a day.

    james> If you take the current date,

    >> as.numeric(chron(dates.="11/23/2004"))
    james> [1] 12745
    >>

    james> you get the value above.  If you change this to
    james> millisecond, you get

    >> as.numeric(chron(dates.="11/23/2004")) * 86400 * 1000
    james> [1] 1.101168e+12
    >>

    james> this value requires 46 bits and since a floating
    james> point number has 54 bits of value, 

no, only 52 bits  (64 = 52+1+12+1) with sign bits for exponent
and mantissa.

    james> it should be enough to give you millisecond resolution and still
    james> maintain the 'date'
"MM" == Martin Maechler <maechler at stat.math.ethz.ch>
    on Tue, 23 Nov 2004 17:04:41 +0100 writes:
"james" == james holtman <james.holtman at convergys.com>
    on Tue, 23 Nov 2004 10:39:04 -0500 writes:
james> You might want to check out 'chron'.  This stores the
    james> time as days and fractions of a day.

    james> If you take the current date,

    >>> as.numeric(chron(dates.="11/23/2004"))
    james> [1] 12745
    >>> 

    james> you get the value above.  If you change this to
    james> millisecond, you get

    >>> as.numeric(chron(dates.="11/23/2004")) * 86400 * 1000
    james> [1] 1.101168e+12
    >>> 

    james> this value requires 46 bits and since a floating
    james> point number has 54 bits of value, 

    MM> no, only 52 bits  (64 = 52+1+12+1) with sign bits for exponent
    MM> and mantissa.

so much on the theme of mathematicians and arithmetic !
   64 = 52 + 1 + 10 + 1

Martin

"james" == james holtman <james.holtman at convergys.com>
    on Tue, 23 Nov 2004 10:39:04 -0500 writes:
   james> You might want to check out 'chron'.  This stores the
   james> time as days and fractions of a day.

   james> If you take the current date,

   >> as.numeric(chron(dates.="11/23/2004"))
   james> [1] 12745
   >>
   james> you get the value above.  If you change this to
   james> millisecond, you get

   >> as.numeric(chron(dates.="11/23/2004")) * 86400 * 1000
   james> [1] 1.101168e+12
   >>
   james> this value requires 46 bits and since a floating
   james> point number has 54 bits of value,

no, only 52 bits  (64 = 52+1+12+1) with sign bits for exponent
and mantissa.
But with an implicit '1' for the first digit in a normalized number.

http://docs.sun.com/source/806-3568/ncg_math.html is one source.  E.g.

`The 52-bit fraction combined with the implicit leading significand bit 
provides 53 bits of precision in double-format normal numbers.'
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595